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Valentin [98]
3 years ago
5

If a steel containing 1.88 wt%C is cooled relatively slowly to room temperature, what is the expected weight fraction of pearlit

e in the as-cooled microstructure? Answer Format: X.XX Unit: unitless (weight fraction)
Physics
1 answer:
Oxana [17]3 years ago
4 0

Answer:

The answer is %pearlite = 0.06%

Explanation:

according to the exercise we have that the percentage is 1.88% C, therefore, the percentage of perlite is equal to:

%pearlite = (B*C)/(A*C) = (2-1.88)/(2-0) = 0.06%

The percentage of cementite is equal to:

%cementite = (1.88-0)/(2-0) = 0.94%

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We already know the length of the pendulum, L=1.38 m, however we need to find its period of oscillation.

We know it makes N=441 oscillations in t=1090 s, therefore its frequency is
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g= \frac{4 \pi^2}{T^2}L =  \frac{4 \pi^2}{(2.47 s)^2} (1.38 m) =8.92 m/s^2
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What is the density (in kg/m3) of a woman who floats in freshwater with 4.92% of her volume above the surface
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